Question

The frequency (f) of a wire oscillating with a length L , in p loops, under a tension T is given by f=p/2L√T/u ,
where u = linear density of the wire. If the error made in determining length, tension and
linear density be 1%, -2% and 4%, then find the percentage error in the calculated frequency.
(1) – 4%
(2) - 2%
(3) -1%
(4) -5%
2eVH​

Answer 1

Given:

Frequency of wire oscillating with a length L , in p loops, under a tension T is given by :

$\boxed{ \sf{f = \dfrac{p}{2L} \sqrt{ \frac{T}{ \mu} } }}$

Error in length, Tension and Linear density is 1% , -2% and 4%.

To find:

Net error in measurement of frequency

Calculation:

$\boxed{ \sf{f = \dfrac{p}{2L} \sqrt{ \frac{T}{ \mu} } }}$

For very small errors of individual components (<6%) , we can say that :

$\sf{ = > \dfrac{\Delta f}{f} = \dfrac{\Delta L}{ L} + \dfrac{1}{2} \bigg( \dfrac{\Delta T}{T} + \dfrac{\Delta \mu}{ \mu} \bigg)}$

$\sf{ = > \dfrac{\Delta f}{f} = 1\% + \dfrac{1}{2} \bigg( 2\% + 4\% \bigg)}$

$\sf{ = > \dfrac{\Delta f}{f} = 1\% + \dfrac{1}{2} \bigg( 6\% \bigg)}$

$\sf{ = > \dfrac{\Delta f}{f} = 1\% + 3\%}$

$\sf{ = > \dfrac{\Delta f}{f} = 4\%}$

So , final answer is:

$\boxed{ \red{ \rm{ \dfrac{\Delta f}{f} = 4\%}}}$

Answer 2

Explanation:

Frequency of wire oscillating with a length L , in p loops, under a tension T is given by :

�

=

�

2

�

�

�

f=

2L

p

μ

T

Error in length, Tension and Linear density is 1% , -2% and 4%.

To find:

Net error in measurement of frequency

Calculation:

�

=

�

2

�

�

�

f=

2L

p

μ

T

For very small errors of individual components (<6%) , we can say that :

=

>

�

�

�

=

�

�

�

+

1

2

(

�

�

�

+

�

�

�

)

=>

f

Δf

=

L

ΔL

+

2

1

(

T

ΔT

+

μ

Δμ

)

=

>

�

�

�

=

1

%

+

1

2

(

2

%

+

4

%

)

=>

f

Δf

=1%+

2

1

(2%+4%)

=

>

�

�

�

=

1

%

+

1

2

(

6

%

)

=>

f

Δf

=1%+

2

1

(6%)

=

>

�

�

�

=

1

%

+

3

%

=>

f

Δf

=1%+3%

=

>

�

�

�

=

4

%

=>

f

Δf

=4%

So , final answer is:

Δ

f

f

=

4

%

f

Δf

=4%